Question

If \( A = \left( \begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix} \right), \, P = \left( \begin{matrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{matrix} \right), \, Q = P^T A P \), then \( PQ^{2014} \) is:

• A. \( \left( \begin{matrix} 1 & 2^{2014}
  0 & 1 \end{matrix} \right) \)
• B. \( \left( \begin{matrix} 1 & 4028
  0 & 1 \end{matrix} \right) \)
• C. \( \left( P^T \right)^{2013} A^{2014} P^{2013} \)
• D. \( P^T A^{2014} P \)

Hint:

When working with matrix powers and transformations, remember that if a matrix is orthogonal (like \( P \) in this case), the product \( P P^T \) simplifies to the identity matrix, which can help simplify calculations.

Answer 1

The Correct Option is B

We are given that: \[ A = \left( \begin{matrix} 1 & 2
0 & 1 \end{matrix} \right), \quad P = \left( \begin{matrix} \cos \theta & \sin \theta
-\sin \theta & \cos \theta \end{matrix} \right), \quad Q = P^T A P \] Now, let's compute \( PQ^{2014} \). Since \( Q = P^T A P \), we can substitute and calculate: \[ Q^{2014} = (P^T A P)^{2014} \] By properties of matrix powers, this simplifies as: \[ Q^{2014} = P^T A^{2014} P \] Thus, \( PQ^{2014} \) is: \[ PQ^{2014} = P \cdot P^T A^{2014} P \] Since \( P P^T = I \) (the identity matrix), the expression simplifies to: \[ PQ^{2014} = A^{2014} \] Now, applying the given value of \( A \): \[ A^{2014} = \left( \begin{matrix} 1 & 2^{2014}
0 & 1 \end{matrix} \right) \] Thus, the correct answer is option (B)